1. Chapter 5
Section 4 and 5

2. Multiplying Polynomials Vertically (cont’d)
CLASSROOM EXAMPLE 3
Multiplying Polynomials Vertically (cont’d)
Find the product.
(5a3 – 6a2 + 2a – 3)(2a – 5)
Solution:
5a3 – 6a2 + 2a – 3
2a – 5
–25a3 + 30a2 – 10a + 15
10a4 – 12a a2 – 6a
10a4 – 37a a2 – 16a + 15
Combine like terms.
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3. Objective 3
Multiply binomials.
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4. Multiply binomials.
When working with polynomials, the products of two binomials occurs repeatedly. There is a shortcut method for finding these products.
First Terms
Outer Terms
Inner Terms
Last Terms
The FOIL method is an extension of the distributive property, and the acronym “FOIL” applies only to multiplying two binomials.
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5. Use the FOIL method to find each product.
CLASSROOM EXAMPLE 4
Using the FOIL Method
Use the FOIL method to find each product.
Solution:
(5r – 3)(2r – 5) F O I L
= (5r)(2r) + (5r)(–5) + (–3)(2r) + (–3)(–5)
= 10r2
– 25r
– 6r
+ 15
= 10r2 – 31r + 15
(4y – z)(2y + 3z) F O I L
= (4y)(2y) + (4y)(3z) + (–z)(2y) + (–z)(3z)
= 8y2 + 12yz – 2yz – 3z2
= 8y2 + 10yz – 3z2
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6. Find the product of the sum and difference of two terms.
Objective 4
Find the product of the sum and difference of two terms.
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7. Product of the Sum and Difference of Two Terms
Find the product of the sum and difference of two terms.
Product of the Sum and Difference of Two Terms
The product of the sum and difference of the two terms x and y is the difference of the squares of the terms.
(x + y)(x – y) = x2 – y2
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8. Multiplying the Sum and Difference of Two Terms
CLASSROOM EXAMPLE 5
Multiplying the Sum and Difference of Two Terms
Find each product.
Solution:
(m + 5)(m – 5)
= m2 – 52
= m2 – 25
(x – 4y)(x + 4y)
= x2 – (4y)2
= x2 – 42y2
= x2 – 16y2
4y2(y + 7)(y – 7)
= 4y2 (y2 – 49)
= 4y4 – 196y2
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9. Find the square of a binomial.
Objective 5
Find the square of a binomial.
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10. Find the square of a binomial. Square of a Binomial
The square of a binomial is the sum of the square of the first term, twice the product of the two terms, and the square of the last term.
(x + y)2 = x2 + 2xy + y2
(x – y)2 = x2 – 2xy + y2
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11. Squaring Binomials Find each product. Solution: (t + 9)2
CLASSROOM EXAMPLE 6
Squaring Binomials
Find each product.
Solution:
(t + 9)2
= t2 + 2• t •
= t t + 81
(2m + 5)2
= (2m)2 + 2(2m)(5) + 52
= 4m m + 25
(3k – 2n)2
= (3k)2 – 2(3k)(2n) + (2n)2
= 9k2 – 12kn + 4n2
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12. Multiplying More Complicated Binomials
CLASSROOM EXAMPLE 7
Multiplying More Complicated Binomials
Find each product.
Solution:
[(x – y) + z][(x – y) – z]
= (x – y)2 – z2
= x2 – 2(x)(y) + y2 – z2
= x2 – 2xy + y2 – z2
(p + 2q)3
= (p + 2q)2(p + 2q)
= (p2 + 4pq + 4q2)(p + 2q)
= p3 + 4p2q + 4pq2 + 2p2q + 8pq2 + 8q3
= p3 + 6p2q + 12pq2 + 8q3
(x + 2)4
= (x + 2)2 (x + 2)2
= (x2 + 4x + 4) (x2 + 4x + 4)
= x4 + 4x3 + 4x2 + 4x x2 + 16x + 4x2 + 16x +16
= x4 + 8x3 + 24x2 + 32x +16
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13. Multiply polynomial functions.
Objective 6
Multiply polynomial functions.
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14. Multiplying Functions
Multiply polynomial functions.
Multiplying Functions
If f (x) and g (x) define functions, then
(fg) (x) = f (x) • g (x) Product function
The domain of the product function is the intersection of the domains of f (x) and g (x).
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15. (fg)(x) = f(x)• g(x). (fg)(2) = 6(2)2 – 13(2) – 5
CLASSROOM EXAMPLE 8
Multiplying Polynomial Functions
For f (x) = 3x + 1 and g (x) = 2x – 5, find (fg) (x) and (fg) (2).
Solution:
(fg)(x) = f(x)• g(x).
= (3x + 1)(2x – 5)
= 6x2 – 15x + 2x – 5
= 6x2 – 13x – 5
(fg)(2) = 6(2)2 – 13(2) – 5
= 24 – 26 – 5
= –7
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16. Chapter 5
Section 5

17. Divide a polynomial by a monomial.
Objective 1
Divide a polynomial by a monomial.
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18. Dividing a Polynomial by a Monomial
Divide a polynomial by a monomial.
Dividing a Polynomial by a Monomial
To divide a polynomial by a monomial, divide each term in the polynomial by the monomial, and then write each quotient in lowest terms.
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19. Dividing a Polynomial by a Monomial (cont’d)
CLASSROOM EXAMPLE 1
Dividing a Polynomial by a Monomial (cont’d)
Divide.
Solution:
Check:
Divisor
Quotient
Original polynomial
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20. Dividing a Polynomial by a Monomial (cont’d)
CLASSROOM EXAMPLE 1
Dividing a Polynomial by a Monomial (cont’d)
Divide.
Solution:
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21. Divide a polynomial by a polynomial of two or more terms.
Objective 2
Divide a polynomial by a polynomial of two or more terms.
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22. Dividing a Polynomial by a Polynomial
CLASSROOM EXAMPLE 2
Dividing a Polynomial by a Polynomial
Divide.
Write the problem as if dividing whole numbers, make sure that both polynomials are written in descending powers of the variables.
Solution:
Divide the first term of 2k2 by the first term of k + 6. Write the result above the division line.
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23. Dividing a Polynomial by a Polynomial (cont’d)
CLASSROOM EXAMPLE 2
Dividing a Polynomial by a Polynomial (cont’d)
+ 5
Multiply and write the result below. 2k(k + 6)
2k2 + 12k 5k
+ 30
Subtract. Do this mentally by changing the signs on 2k2 + 12k and adding.
5k + 30
Bring down 30 and continue dividing 5k by k.
Subtract.
You can check the result, 2k + 5, by multiplying k + 6 and 2k + 5.
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24. Dividing a Polynomial with a Missing Term
CLASSROOM EXAMPLE 3
Dividing a Polynomial with a Missing Term
Divide 4x3 + 3x – 8 by x + 2.
Write the polynomials in descending order of the powers of the variables.
Add a term with 0 coefficient as a placeholder for the missing x2 term.
Solution:
Missing term
Remember to include as part of the answer. Don’t forget to insert a plus sign between the polynomial quotient and this fraction.
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25. Dividing a Polynomial with a Missing Term (cont’d)
CLASSROOM EXAMPLE 3
Dividing a Polynomial with a Missing Term (cont’d)
Start with
4x2
 8x
+ 19
Subtract by mentally by changing the signs on 4x3 + 8x2 and adding.
4x3 + 8x2
Bring down the next term.
Next,
8x2
+ 3x
8x2 – 16x
Multiply then subtract.
19x
 8
Bring down the next term.
19x + 38
19x/19 = 19
Remainder
–46
Multiply then subtract.
The solution is: and you can check the result by multiplying.
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26. Dividing a Polynomial with a Missing Term
CLASSROOM EXAMPLE 4
Dividing a Polynomial with a Missing Term
Divide 4m4 – 23m3 + 16m2 – 4m – 1 by m2 – 5m.
Write the polynomial m2 – 5m as m2 – 5m + 0.
Solution:
Missing term
Since the missing term is the last term it does not need to be written.
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27. Dividing a Polynomial with a Missing Term (cont’d)
CLASSROOM EXAMPLE 4
Dividing a Polynomial with a Missing Term (cont’d)
4m2
 3m
+ 1
4m4 – 20m3
3m3
+ 16m2
3m m2 m2
– 4m
m2 – 5m
m – 1
Remainder
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28. Finding a Quotient with a Fractional Coefficient
CLASSROOM EXAMPLE 5
Finding a Quotient with a Fractional Coefficient
Divide 8x3 + 21x2 – 2x – 24 by 4x + 8.
Solution:
2x2 +
 3
8x3 + 16x2
5x2
 2x
5x2 + 10x
12x
– 24
The solution is:
–12x – 24
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29. Divide polynomial functions.
Objective 3
Divide polynomial functions.
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30. Divide polynomial functions. Dividing Functions
If f (x) and g (x) define functions, then
Quotient function
The domain of the quotient function is the intersection of the domains of f (x) and g (x), excluding any values of x for which g (x) = 0.
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31. Dividing Polynomial Functions
CLASSROOM EXAMPLE 6
Dividing Polynomial Functions
For f(x) = 2x2 + 17x + 30 and g(x) = 2x + 5, find
Solution:
From previous Example 2, we conclude that (f/g)(x) = x + 6, provided the denominator 2x + 5, is not equal to zero.
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